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What are the domain and range of the function $y = 2 ext{cos}^{-1}(2x) + 2 ext{sin}^{-1}(2y)$? - HSC - SSCE Mathematics Extension 1 - Question 4 - 2024 - Paper 1

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What-are-the-domain-and-range-of-the-function-$y-=-2--ext{cos}^{-1}(2x)-+-2--ext{sin}^{-1}(2y)$?-HSC-SSCE Mathematics Extension 1-Question 4-2024-Paper 1.png

What are the domain and range of the function $y = 2 ext{cos}^{-1}(2x) + 2 ext{sin}^{-1}(2y)$?

Worked Solution & Example Answer:What are the domain and range of the function $y = 2 ext{cos}^{-1}(2x) + 2 ext{sin}^{-1}(2y)$? - HSC - SSCE Mathematics Extension 1 - Question 4 - 2024 - Paper 1

Step 1

Domain: Determine the valid input values for $x$

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Answer

Given the inverse functions involved, the argument of extcos1 ext{cos}^{-1} and extsin1 ext{sin}^{-1} must fall within the ranges of 12x1-1 \leq 2x \leq 1 for extcos1 ext{cos}^{-1} and 12y1-1 \leq 2y \leq 1 for extsin1 ext{sin}^{-1}. Setting these inequalities gives us:

  1. From 12x1-1 \leq 2x \leq 1:

    • Dividing everything by 2 results in: 0.5x0.5-0.5 \leq x \leq 0.5.
  2. Similarly, for yy:

    • 12y1-1 \leq 2y \leq 1 gives: 0.5y0.5-0.5 \leq y \leq 0.5.

Thus, considering both variables, the domain can be specified as: [0.5,0.5][-0.5, 0.5].

Step 2

Range: Determine the output values of the function

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Answer

The range of the function is influenced by the inverse trigonometric functions:

  • For extcos1(x) ext{cos}^{-1}(x), the output is limited to [0,pi][0, \\pi]. Since we have 2extcos1(2x)2 ext{cos}^{-1}(2x), this stretches the output to [0,2pi][0, 2\\pi].
  • For extsin1(y) ext{sin}^{-1}(y), the output is [fracπ2,fracπ2][-\\frac{\pi}{2}, \\frac{\pi}{2}], and thus, 2extsin1(2y)2 ext{sin}^{-1}(2y) alters the range to [π,π][-\pi, \pi].

Considering the combination of these outputs, we can conclude that the range of the overall function is: {π}\{\pi\}.

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